# Questions tagged [cauchy-product]

For questions about the Cauchy product, the discrete convolution of two sequences.

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### An example: a convergence series, a divergent series, whose Cauchy product is convergent.

How to find an example: a convergence series $\sum a_n$, a divergent series $\sum b_n$, whose Cauchy product $\sum c_n$ with $c_n=\sum_{i+j=n}a_ib_j$ is convergent? Is there a simple example?
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### The Cauchy Product : Calculate the coefficients, $c_n$, in the product.

I found this definition to be different from other ones and the answer makes me feel weird. If $$A_{N}(x)=\sum_{n=0}^{n=N}a_{n}x^n$$ and $$B_{N}(x)=\sum_{n=0}^{n=N}b_{n}x^n,$$ calculate the ...
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### Are the unconditionally convergent series, with terms in a Banach algebra, closed under the Cauchy product?

We have a Banach algebra $\mathbb L$, and two sequences $(A_0,A_1,A_2,\cdots),\;(B_0,B_1,B_2,\cdots)\in\mathbb L^{\mathbb N}$, for which the sums $\sum_{n\in\mathbb N}A_n,\;\sum_{n\in\mathbb N}B_n$ ...
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### The necessity of absolute convergence in the convergence of the Cauchy product of series?

The Mertens' theorem claims that Suppose $\sum_{n=0}^\infty a_n,\sum_{n=0}^\infty b_n$ are two convergent series of complex numbers, convergent to $A,\beta$ respectively. If $\sum_na_n$ converges ...
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### How to show that $\sum_{k=0}^{\infty}{(k+1)z^k}=\frac{1}{(1-z)^2}$ for $z\in \mathbb{C}:|z|<1$

How to show that $\sum_{k=0}^{\infty}{(k+1)z^k}=\frac{1}{(1-z)^2}$ for $z\in \mathbb{C}:|z|<1$ I believe that I have to use the cauchy product? But how do transform the expression to be a product ...
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### Product of Two Summations with Different Upper Limits

I'm trying to multiply two different finite summations with different upper limits. I've tried Cauchy Product but i think it's valid for same upper limits. I also tried to split the summation. Any ...
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### Find the power series of $f(x)=\frac{x}{x+1}$ in $x_0=0$

In 1st Semester Calculus book I found an exercise that asks me to find the above power series of the function at the point $x_0 = 0$ using the geometric series formula and the Cauchy-Product. So far I'...
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### Cauchy product of power series with $x^{2n}$

I am trying to rewrite a function $y(x)=\frac{1}{1+x+x^2+x^3}$ as a series. I used geometric series and got to two power series that I need to Cauchy product, however, one of them has $x^{2n}$ and I ...
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### Infinitely Nested Infinite Series / Infinite Composition of Series

Is there any documentation about a series like this? What is it called? Does it have a value? I have tried several searches and couldn't find anything close to this. Any guidance would be ...
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### Cauchy product for the reciprocal of the polynomial $x - 2x^2 + 3x^3 - 4x^4$

I have come across some Laurent series in which the denominator of a fraction contains a power series. Looking around I came across this Calculate Laurent series for $1/ \sin(z)$ which suggests that ...
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### Power series of $\frac{1}{(1-z)^2}$

I want to show, that the following is true for every $z\in C$ with $|z|<1$: $$\frac{1}{(1-z)^2} =\sum_{k=1}^\infty kz^{k-1}$$ I think there is a way with the Cauchy-Product
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### Decomposition in product

I start my question with an example that I did. Give the following series $R(x):=\sum_{d=1}^{\infty}(\sum_{l=0}^{d-1} (-1)^{l}\frac{1}{ \ \ell!(d-1-\ell)!})q_{1}^{d} x^d$. I could decompose this as a ...
How to find the sum of $$\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left\{\sum_{k=1}^{2n+1}\frac{(-1)^k} k \right\}$$ $$\begin{array}\\ \frac{1}{1}&\times&(-\frac{1}{1}) &+&\ \ \\ (-\... 4answers 185 views ### Find limit of sum I suspect that \lim_{n \to \infty} \sum_{k = 0}^{n - 1}\frac{k}{a^k(n - k)} = 0 for a > 1. I know that this product represents the taylor coefficients of \frac{-ax\ln(1 - x)}{(a - x)^2} by ... 3answers 116 views ### The series \sum_{k=1}^\infty\sum_{n=1}^k k^{-3}/(2n-1)$$\sum_{k=1}^\infty\sum_{n=1}^k\frac{1}{(2n-1)k^3} Can anyone help me find this series? I tried to use Cauchy product but I don't know how I can complete it.
I was looking at the counterexample $a_n = b_n = (-1)^n/\sqrt{n}$ where $\sum a_n$ and $\sum b_n$ converge but the Cauchy product $\sum_{k=0}^\infty c_k = \sum_{k=0}^\infty \sum_{j=0}^{k}a_j b_{k-j}$ ...